Question

Determine whether x=3/2 and x=-4/3 are the solutions of the equation 6x2 — x — 12 = 0 or not.

Answer 1

Question:

Determine whether x=3/2 and x=-4/3 are the solutions of the equation 6x² - x - 12 = 0 or not.

Answer:

x = 3/2 and x = -4/3 are the solutions of the given equation.

Note:

• The possible values of unknown (variable) for which the equation is satisfied are called its solutions or roots .

• If x = a is a solution of any equation in x , then it must satisfy the given equation otherwise it's not a solution (root) of the equation.

Solution:

The given equation is : 6x² - x - 12 = 0 ------(1)

Let's check whether x = 3/2 is a solution of eq-(1) or not .

Putting x = 3/2 in eq-(1) , we have ;

=> 6x² - x - 12 = 0

=> 6(3/2)² - 3/2 - 12 = 0

=> 6•(9/4) - 3/2 - 12 = 0

=> 54/4 - 3/2 - 12 = 0

=> 27/2 - 3/2 - 12 = 0

=> (27 - 3 - 24)/2 = 0

=> 0/2 = 0

=> 0 = 0 (which is true)

Since , eq-(1) is satisfied by x = 3/2 , thus x = 3/2 is a solution of eq-(1) .

Now,

Let's check whether x = -4/3 is a solution of eq-(1) or not .

Putting x = -4/3 in eq-(1) , we have ;

=> 6x² - x - 12 = 0

=> 6(-4/3)² - (-4/3) - 12 = 0

=> 6•(16/9) + 4/3 - 12 = 0

=> 32/3 + 4/3 - 12 = 0

=> (32 + 4 - 36)/3 = 0

=> 0/3 = 0

=> 0 = 0 (which is true)

Since , eq-(1) is satisfied by x = -4/3 , thus x = -4/3 is a solution of eq-(1) .

Answer 2

$\huge{\boxed{\red{Answer}}}$

$\large{\underline{\pink{Required\;to\;find}}}$

$Whether\;\dfrac{3}{2}\;and\;\dfrac{-4}{3}\;are\;roots\;of\;equation\\ 6x^{2}-x-12=0$

$\large{\underline{\pink{Verifying\;\dfrac{3}{2}\;as\;a\;root}}}$

• Substituting $\dfrac{3}{2}$ in the equation
• $6x^{2}-x-12=0$

• $6{(\dfrac{3}{2})}^{2}-\dfrac{3}{2}-12=0$

• $\dfrac{27-3-24}{2}=0$

• 0=0

$\boxed{\green{Therefore\;\dfrac{3}{2}\;is\;a\;root}}$

$\large{\underline{\pink{Verifying\;\dfrac{-4}{3}\;as\;a\;root}}}$

• Substituting $\dfrac{-4}{3}$ in the equation
• $6x^{2}-x-12=0$
• $6(\dfrac{16}{3})-\dfrac{-4}{3}-12=0$

• $\dfrac{32+4-36}{3}=0$

• 0=0

$\boxed{\green{Therefore\;\dfrac{-4}{3}\;is\;a\;root}}$

$\boxed{\red{Therefore\;\dfrac{3}{2}\;and\;\dfrac{-4}{3}\;are\;roots\;of\;given\;equation}}$

$\huge{\boxed{\blue{NOTE}}}$

• If 'a' is a root of a given equation then it must satisfy the given equation
• I have used the above statement to verify -3 as root of given equation